Question

prove that the bisector of the vertical angle of an isosceles triangle bisects the base at right angle



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Answer 1

Consider PQR is an isosceles Triangle

Such that

PQ = PR -----( Opposite sides of Isosceles triangle are equal)

=> Pl is the bisector of ∠P.

To prove :

∠PLQ = ∠PLR = 90° & QL = LR

To Proof:-

In ΔPLQ and ΔPLR

• PQ = PR -----(given)

• PL = PL -----(common)

• ∠QPL = ∠RPL -----( PL is the bisector of ∠P)

• ΔPLQ = ΔPLR ----( SAS congruence criterion)

=> QL = LR ---(by cpct)

=> ∠PLQ + ∠PLR = 180° -----( linear pair)

=> 2∠PLQ = 180°

=> ∠PLQ = 180° / 2 = 90°

∴ ∠PLQ = ∠PLR = 90°

Thus, ∠PLQ = ∠PLR = 90° and QL = LR

Hence, the bisector of the verticle angle an isosceles triangle bisects the base at right angle.

Hence Proved!

Answer 2

Answer:

Consider ΔABC and AM is the bisector of ∠A and AM bisects the base BC so BM = CM

Extend segment AM to D such that AM = MD and join points C and D to from ΔDMC as shown

To prove ΔABC is isosceles we have to prove that AB = AC

Consider ΔAMB and ΔDMC

AM = MD … construction

∠AMB = ∠DMC … vertically opposite angles

BM = MC … AM bisects BC given